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dezoksy [38]
3 years ago
15

Explain how the perfume atomizer works under Bernoulli's principle

Physics
1 answer:
ss7ja [257]3 years ago
4 0

Answer:

In an atomizer, or perfume sprayer, you squeeze a rubber bulb to squirt air through a tube. Because of the Bernoulli principle, the air rushing through the tube has a lower pressure than the surrounding atmosphere. ... The perfume is pushed out of the tube and sprays into the air as a fine mist.

Explanation:

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an ice sheet 5m thick covers a lake that is 20m deep. at what is the temperature of the water at the bottom of the lake?
muminat

Answer:

4°C

Explanation:

Water is densest at 4°C.  Since dense water sinks, the bottom of the lake will be 4°C.

6 0
3 years ago
Graphs are representations of equations.<br> A. True<br> B. False
zhuklara [117]

Answer:

Hi there!

The answer is: A. True

6 0
3 years ago
Read 2 more answers
I need help with part D
natka813 [3]
1).  The little projectile is affected by friction all the way through the block.
Friction robs some kinetic energy.

2).  The block is affected by friction as it scrapes along the top of the post.
Friction robs some kinetic energy.

3).  The block is also affected by friction with the air (air resistance) as it
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8 0
3 years ago
A cyclical heat engine, operating between temperatures of 450º C and 150º C produces 4.00 MJ of work on a heat transfer of 5.00
gogolik [260]

Answer:

(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%

Explanation:

Using the formula that relate heat and work from the thermodynamic theory as:W=Q=Q_{in}-Q_{out} solving to Q_out we get:Q_{out}=Q_{in}-W=5(MJ)-4(MJ)=1(MJ) this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: n=1-\frac{T_{Low} }{T_{High}} where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:n=1-\frac{423.15}{723.15} =0.415=41.5%

5 0
3 years ago
The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0
katen-ka-za [31]

Answer:

a)  F = 2.66 10⁴ N, b)   h = 1.55 m

Explanation:

For this fluid exercise we use that the pressure at the tap point is

Exterior

          P₂ = P₀ = 1.01 105 Pa

inside

         P₁ = P₀ + ρ g h

the liquid is water with a density of ρ=1000 km / m³

         P₁ = 0.85   1.01 10⁵ + 1000   9.8  5

         P₁ = 85850 + 49000

         P₁ = 1.3485 10⁵ Pa

the net force is

         ΔP = P₁- P₂

         Δp = 1.3485 10⁵ - 1.01 10⁵

         ΔP = 3.385 10⁴ Pa

Let's use the definition of pressure

         P = Fe / A

         F = P A

the area of ​​a circle is

         A = pi r² = [i d ^ 2/4

let's reduce the units to the SI system

         d = 100 cm (1 m / 100 cm) = 1 m

         F = 3.385 104 pi / 4 (1) ²

         F = 2.66 10⁴ N

b) the height for which the pressures are in equilibrium is

        P₁ = P₂

        0.85 P₀ + ρ g h = P₀

        h = \frac{P_o ( 1-0.850)}{\rho \ g}

        h = \frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}

        h = 1.55 m

4 0
3 years ago
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